A charged particle of charge $e$ and mass $m$ is moving in an electric field $\vec{E}$ and magnetic field $\vec{B}$. Construct dimensionless quantities and quantities of dimension $[T]^{-1}$.

A charged particle of charge e and mass m is moving in an electric field vecE and magnetic field vecB. Constru

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Concept understanding — Charged Particle in Magnetic Field

Charged Particle in a Magnetic Field

When a charged particle moves through a magnetic field, the field grabs it sideways. Unlike an electric field, which can speed a charge up or slow it down, a magnetic field only bends the path — it never changes the particle's speed. Understanding why leads directly to circular and helical motion, the basis of cyclotrons, mass spectrometers and the aurora.

The force: always sideways

A particle of charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ feels the magnetic (Lorentz) force:

\vec{F} = q\,(\vec{v} \times \vec{B})

Because of the cross product, $\vec{F}$ is perpendicular to both $\vec{v}$ and $\vec{B}$ . Its magnitude is

F = |q|\,v\,B\sin\theta

where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$ .

Important

Since $\vec{F} \perp \vec{v}$ , the force does no work: $\vec{F}\cdot\vec{v} = 0$ . Therefore the kinetic energy and the speed stay constant — the field only changes the direction of motion, never the magnitude.

Case 1: velocity perpendicular to the field → a circle

If $\vec{v} \perp \vec{B}$ ( $\theta = 90^\circ$ ), the force $F = qvB$ stays constant in size and always points toward one central point. That is exactly the condition for uniform circular motion, with the magnetic force acting as the centripetal force:

qvB = \frac{mv^2}{r}

Solving for the radius:

r = \frac{mv}{qB}

The time period of one revolution is

T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}

Note

The period $T$ (and the frequency $f = qB/2\pi m$ , the cyclotron frequency) does not depend on the speed or the radius. A faster particle traces a bigger circle but takes exactly the same time per loop. This speed-independence is what makes the cyclotron work.

Case 2: velocity at an angle → a helix

If $\vec{v}$ makes an angle $\theta$ with $\vec{B}$ , split it into two parts:

Perpendicular component $v_\perp = v\sin\theta$ — feels the magnetic force and drives circular motion of radius $r = \dfrac{mv_\perp}{qB}$ .
Parallel component $v_\parallel = v\cos\theta$ — feels no force (since $\vec{v_\parallel}\times\vec{B}=0$ ) and carries the particle steadily along the field line.

Combining a circle with a steady drift gives a helix. The distance advanced along the field in one full turn is the pitch:

p = v_\parallel\,T = v\cos\theta \cdot \frac{2\pi m}{qB}

A quick example

An electron ( $m = 9.1\times10^{-31}\ \text{kg}$ , $q = 1.6\times10^{-19}\ \text{C}$ ) enters a $0.02\ \text{T}$ field at $10^{6}\ \text{m/s}$ , perpendicular to $\vec{B}$ :

$r = \frac{mv}{qB} = \frac{(9.1\times10^{-31})(10^{6})}{(1.6\times10^{-19})(0.02)} \approx 2.8\times10^{-4}\ \text{m}$ …

Why this formula?

Charged Particle in a Magnetic Field — Why the Key Formulas Hold

Let's build this from first principles. The core idea is that a magnetic field exerts a force only on a moving charge, and that force is always perpendicular to both the velocity and the field.

1. The Fundamental Force Law: Lorentz Force

The starting point is the Lorentz force for a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ :

\vec{F}_m = q (\vec{v} \times \vec{B})

Why this form?

Cross product $\vec{v} \times \vec{B}$ means the force is perpendicular to both $\vec{v}$ and $\vec{B}$ .
Magnitude: $F_m = |q| v B \sin\theta$ , where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$ .
Direction: given by the right-hand rule (for positive $q$ ).

Key insight: Because $\vec{F}_m \perp \vec{v}$ , the magnetic force does no work — it changes only the direction of velocity, not its speed.

2. Circular Motion in a Uniform Magnetic Field

Consider a charge $q$ moving with speed $v$ perpendicular to a uniform $\vec{B}$ (so $\theta = 90^\circ$ , $\sin\theta = 1$ ).

Step 1: Force provides centripetal acceleration

The magnetic force is the only radial force:

F_m = q v B

This must equal the centripetal force required for circular motion:

F_c = \frac{m v^2}{r}

Step 2: Equate and solve for $r$

q v B = \frac{m v^2}{r}

Cancel one $v$ (assuming $v \neq 0$ ):

q B = \frac{m v}{r}

Thus:

\boxed{r = \frac{m v}{q B}}

This is the radius of the circular path (cyclotron radius).

Why this makes sense:

Larger mass $m$ → harder to turn → larger $r$
Larger charge $q$ or stronger $B$ → stronger force → tighter turn → smaller $r$
Faster speed $v$ → more momentum → larger $r$

3. Angular Frequency (Cyclotron Frequency)

From the circular motion relation:

\omega = \frac{v}{r}

Substitute $r = \frac{m v}{q B}$ :

\omega = \frac{v}{\frac{m v}{q B}} = \frac{q B}{m}

Thus:

\boxed{\omega_c = \frac{q B}{m}}

Why this is remarkable:

$\omega_c$ is independent of speed $v$ — all particles with same $q/m$ have the same angular frequency, regardless of how fast they move.
This is the principle behind cyclotrons (particle accelerators).

4. General Motion: Helical Path …

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The key idea is to use dimensional analysis to combine $e$ , $m$ , $E$ , and $B$ into quantities with dimensions of $[T]^{-1}$ (frequency). The cyclotron frequency $\omega_c = eB/m$ is one such quantity, and another is $\sqrt{eE/md}$ where $d$ is a length scale — but without a length scale, only $eB/m$ is purely $[T]^{-1}$ from the given parameters.

Why Magnetic Force Balance?

When a charged particle moves through electric and magnetic fields, two forces act on it: the electric force $\vec{F}_E = e\vec{E}$ and the magnetic force $\vec{F}_B = e(\vec{v} \times \vec{B})$ . The magnetic force depends on velocity, which introduces a natural timescale into the problem.

The most famous frequency that emerges is the cyclotron frequency — the rate at which a charged particle gyrates in a uniform magnetic field. This comes from equating the magnetic force to the centripetal force: $evB = m v^2 / r$ , giving $\omega = v/r = eB/m$ . Notice that this frequency depends only on $e$ , $B$ , and $m$ — not on $v$ or $r$ individually.

But the problem also includes an electric field $E$ . Can we construct another frequency involving $E$ ? Let's see.

Step-by-step construction

1. Identify the dimensions of each quantity

First, let's write down the dimensions of everything we have:

Charge $e$ : $[e] = [\text{current} \times \text{time}] = [I T]$
Mass $m$ : $[m] = [M]$
Electric field $E$ : $[E] = [\text{force/charge}] = [M L T^{-2}]/[I T] = [M L T^{-3} I^{-1}]$
Magnetic field $B$ : $[B] = [\text{force/(charge × velocity)}] = [M L T^{-2}]/[I T \cdot L T^{-1}] = [M T^{-2} I^{-1}]$

We want quantities with dimension $[T]^{-1}$ , i.e., frequency.

2. The obvious candidate: $eB/m$

Take the combination $eB/m$ :

$[eB] = [I T] \cdot [M T^{-2} I^{-1}] = [M T^{-1}]$
Dividing by $m$ : $[eB/m] = [M T^{-1}]/[M] = [T^{-1}]$

So $\omega_c = \frac{eB}{m}$ is a frequency. This is the cyclotron frequency — the angular frequency of circular motion in a pure magnetic field.

\omega_c = \frac{eB}{m}

3. Can we get another frequency involving $E$ ?

Try $eE/m$ :

$[eE] = [I T] \cdot [M L T^{-3} I^{-1}] = [M L T^{-2}]$
$[eE/m] = [M L T^{-2}]/[M] = [L T^{-2}]$

This has dimensions of acceleration, not frequency. To get $[T^{-1}]$ , we'd need to divide by a velocity or multiply by a time — but we don't have those as independent parameters.

What about $\sqrt{eE/md}$ ? If we introduce a length $d$ , then:

$[eE/md] = [L T^{-2}]/[L] = [T^{-2}]$
Taking square root gives $[T^{-1}]$

But the problem only gives $e$ , $m$ , $E$ , and $B$ — no length scale. So without additional parameters, $eE/m$ alone cannot yield a frequency.

Watch out

A common mistake is to think $eE/m$ gives a frequency. It doesn't — it gives acceleration. You need a length or velocity to convert it into a frequency.

4. The mixed combination: $E/B$

What about $E/B$ ?

$[E/B] = [M L T^{-3} I^{-1}]/[M T^{-2} I^{-1}] = [L T^{-1}]$

This is a velocity — the drift velocity $v_d = E/B$ for crossed fields. If we then divide by a length, we get frequency. But again, no length is given.

5. What dimensionless quantities can we form?

A dimensionless quantity requires the product of powers of $e$ , $m$ , $E$ , $B$ to have dimension $[M^0 L^0 T^0 I^0]$ . Let's try:

$e^a m^b E^c B^d$ has dimension:

$[I^a T^a] [M^b] [M^c L^c T^{-3c} I^{-c}] [M^d T^{-2d} I^{-d}]$

Collecting: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 25 on this concept.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$1.66\times10^{-27}\ kg$
›Reveal solutionSolution
This tests the magnetic Lorentz force on a moving charge, after first extracting speed from kinetic energy; the force is $\approx7.8\times10^{-12}\ N$ .

Concept and Intuition

The magnetic force on a charge moving perpendicular to a field is $F = qvB$ (maximum, since $\sin90^\circ=1$ ). The only extra step here is that the speed isn't given directly — it must be extracted from the kinetic energy using $KE = \frac{1}{2}mv^2$ , converting MeV to joules first.

Step-by-Step Solution

Convert energy: $KE = 2\ MeV = 2\times1.6\times10^{-13}\ J = 3.2\times10^{-13}\ J$ .

Solve for speed: $v = \sqrt{\dfrac{2KE}{m}} = \sqrt{\dfrac{2\times3.2\times10^{-13}}{1.66\times10^{-27}}} = \sqrt{3.855\times10^{14}} \approx 1.9635\times10^7\ m/s$ .

Since the proton moves perpendicular to $B$ , the force is maximum: $F = qvB$ . …
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AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$E = 2\times10^3\hat{i}\ N\,C^{-1}$
›Reveal solutionSolution
This is the classic velocity-selector condition — zero net force means the electric and magnetic forces exactly cancel, fixing both the speed ( $v=E/B$ ) and the direction of $\vec v$ .

Concept and Intuition

A charged particle experiences zero acceleration only when the total force is zero. Here the forces are electric ( $q\vec E$ ) and magnetic ( $q\vec v\times\vec B$ ). For these to cancel exactly (for any nonzero charge), their magnitudes must match and their directions must be opposite. This is exactly the working principle of a velocity selector — only particles with a specific speed $v=E/B$ pass through undeflected, regardless of charge magnitude.

Step-by-Step Solution

Balance condition: $q\vec E + q(\vec v\times \vec B) = 0 \Rightarrow \vec v \times \vec B = -\vec E$ .

Magnitude: $vB = E \Rightarrow v = \dfrac{E}{B} = \dfrac{2\times10^3}{5\times10^{-2}} = 4\times10^4\,\text{m/s}$ . …
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AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$qvB$
›Reveal solutionSolution
The magnetic force on a moving charge is $F=qvB\sin\theta$ ; when velocity is parallel to $\vec B$ , the angle between them is zero, so the force vanishes. Answer: zero.

Concept and Intuition

The magnetic Lorentz force is $\vec F = q\vec v \times \vec B$ , whose magnitude is $qvB\sin\theta$ , where $\theta$ is the angle between velocity and field. This is a cross product, and cross products of parallel (or anti-parallel) vectors are always zero — there is no component of velocity perpendicular to $\vec B$ to generate a deflecting force.

Step-by-Step Solution

Force law: $F = qvB\sin\theta$ .

"Moving parallel to the field" means the angle between $\vec v$ and $\vec B$ is $\theta = 0^\circ$ . …
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AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$6\ \mathrm{km\,s^{-1}}$
›Reveal solutionSolution
This tests the velocity selector principle: crossed electric and magnetic fields let through only one speed. Answer: $B=\dfrac{1}{15}$ T.

Concept and Intuition

In a velocity selector, an electric field and a magnetic field are arranged perpendicular to each other and to the ion's velocity, so the electric force $qE$ and magnetic force $qvB$ act in opposite directions. Only ions moving at the exact speed where these forces balance pass through undeflected: $qE=qvB \Rightarrow v=\dfrac{E}{B}$ .

Step-by-Step Solution

Balance condition: $v=\dfrac{E}{B} \Rightarrow B=\dfrac{E}{v}$ .

Convert velocity: $v=6\ \text{km/s}=6000\ \text{m/s}$ . …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$V_1$
›Reveal solutionSolution
The maximum magnetic force on a charge scales with its speed, and speed scales with the square root of the accelerating voltage; a 4-fold force increase means a 16-fold voltage increase, so $V_1:V_2=1:16$ .

Concept and Intuition

When a charge is accelerated from rest through a potential difference $V$ , all the electrical work goes into kinetic energy: $qV=\tfrac12 mv^2$ , so $v=\sqrt{2qV/m}$ — speed grows only as the square root of the voltage. Once in a uniform magnetic field, the maximum magnetic force (when velocity is perpendicular to $B$ ) is $F_{max}=qvB$ , which is directly proportional to $v$ , and hence to $\sqrt{V}$ .

Step-by-Step Solution

Speed after acceleration through $V_1$ : $v_1=\sqrt{2qV_1/m}$ , giving force $F=qv_1B$ .

Speed after acceleration through $V_2$ : $v_2=\sqrt{2qV_2/m}$ , giving force $4F=qv_2B$ . …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$= 1.67 \times 10^{-27}$
›Reveal solutionSolution
This tests converting kinetic energy to speed and then applying the magnetic Lorentz force $F=qvB$ for a charge moving perpendicular to $B$ ; the answer is $64\times10^{-12}$ N.

Concept and Intuition

A charged particle moving at right angles to a uniform magnetic field feels a force $F=qvB\sin\theta=qvB$ (since $\theta=90^\circ$ ). To use this we first need the proton's speed, which we get from its kinetic energy.

Step-by-Step Solution

Convert kinetic energy to joules using the proton charge as the eV-to-J factor: $KE=8.35\times1.6\times10^{-13}\,\text{J}=1.336\times10^{-12}$ J.

From $KE=\tfrac12 mv^2$ : $v=\sqrt{\dfrac{2KE}{m}}=\sqrt{\dfrac{2\times1.336\times10^{-12}}{1.67\times10^{-27}}}=\sqrt{1.60\times10^{15}}\approx4.0\times10^{7}\ \text{m/s}$ . …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$(2\bar{i} + 3\bar{j})\ ms^{-1}$
›Reveal solutionSolution
Computing $\vec v\times\vec B$ and adding $\vec E$ gives the net field vector $(12,0,6)$ V/m; its magnitude times $e$ gives $2.15\times10^{-18}$ N at $\cos^{-1}(2/\sqrt5)$ from the x-axis.

Concept and Intuition

The Lorentz force is $\vec F=q(\vec E+\vec v\times\vec B)$ . We compute the cross product component-wise, add the electric field, then find magnitude and direction of the resultant.

Step-by-Step Solution

$\vec v\times\vec B$ with $\vec v=(2,3,0)$ , $\vec B=(0,2,3)$ :

$\vec v\times\vec B = (3\cdot3-0\cdot2,\ 0\cdot0-2\cdot3,\ 2\cdot2-3\cdot0) = (9,-6,4)$

Add $\vec E=(3,6,2)$ : $\vec E+\vec v\times\vec B = (12,0,6)$ .

Magnitude: $|\vec F\text{-field}| = \sqrt{12^2+0^2+6^2}=\sqrt{180}=6\sqrt5$ .

Force: $F = e\times6\sqrt5 = 1.6\times10^{-19}\times13.42 \approx 2.15\times10^{-18}$ N. …
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AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
$Q. A proton and an alpha particle moving with energies in the ratio 1:4 enter a uniform magnetic field of 3T at right angles to the direction of magnetic field. The ratio of the magnetic forces acting on the proton and the alpha particle is (A) 1:2 (B) 1:4 (C) 2:3 (D) 1:3$
›Reveal solutionSolution
Express the magnetic force in terms of kinetic energy (not just charge and mass separately), then substitute the known charge/mass/energy ratios for proton and alpha. Answer: 1:2.

Concept and Intuition

The magnetic force on a charged particle moving perpendicular to a field is $F = qvB$ . Since we're given kinetic energies rather than speeds, it helps to write $v = \sqrt{2KE/m}$ and substitute, so that $F = qB\sqrt{2KE/m}$ .

Step-by-Step Solution

Proton: charge $q_p = e$ , mass $m_p$ , kinetic energy $K$ .

Alpha particle: charge $q_\alpha = 2e$ , mass $4m_p$ , kinetic energy $4K$ (given ratio 1:4).

$F_p = eB\sqrt{\dfrac{2K}{m_p}}$ .

$F_\alpha = 2eB\sqrt{\dfrac{2(4K)}{4m_p}} = 2eB\sqrt{\dfrac{2K}{m_p}}$ . …
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AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
$8 \times 10^7\ \mathrm{C\,kg^{-1}}$
›Reveal solutionSolution
Cyclotron angular frequency depends only on the specific charge (charge-to-mass ratio) and the field strength. Answer: $32\times10^4$ rad/s.

Concept and Intuition

When a charged particle moves perpendicular to a uniform magnetic field, it moves in a circle with angular velocity $\omega = \dfrac{qB}{m}$ . This is independent of the particle's speed or the radius of its path — it depends only on the specific charge $q/m$ and the field $B$ .

Step-by-Step Solution

Specific charge: $q/m = 8\times10^7\ \mathrm{C\,kg^{-1}}$ .

Field: $B = 4\ \mathrm{mT} = 4\times10^{-3}\ \mathrm{T}$ . …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$V_1$
›Reveal solutionSolution
Speed after acceleration scales as $\sqrt V$ , and magnetic force scales linearly with speed, so force scales as $\sqrt V$ ; doubling the force means quadrupling the potential difference.

Concept and Intuition

Accelerating the electron through a potential difference converts electrical PE into KE: $qV = \tfrac12 mv^2$ , so $v=\sqrt{2qV/m}$ — speed grows only as the square root of the accelerating voltage. Once moving in the magnetic field, the magnetic (Lorentz) force is $F=qvB$ , which is directly proportional to speed $v$ (with $q$ , $B$ fixed). Chaining these two proportionalities, $F\propto\sqrt V$ .

Step-by-Step Solution

From energy: $v = \sqrt{\dfrac{2qV}{m}} \Rightarrow v\propto\sqrt V$ .

Magnetic force: $F = qvB \propto v \propto \sqrt V$ .

Given $F_2 = 2F_1$ : $\dfrac{F_2}{F_1}=\sqrt{\dfrac{V_2}{V_1}}=2$ . …
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AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$2 \times 10^5\ \mathrm{ms^{-1}}$
›Reveal solutionSolution
Computing $\vec v \times \vec B$ for the given vectors gives a magnitude of $5v$ ; multiplying by the electron charge yields a magnetic force of $1.6\times10^{-13}$ N.

Concept and Intuition

A charged particle moving through a magnetic field experiences the Lorentz magnetic force $\vec F = q\vec v \times \vec B$ , whose magnitude depends only on the component of velocity perpendicular to $\vec B$ . Direct vector cross-product computation is the cleanest way to handle a $\vec B$ with multiple non-zero components.

Step-by-Step Solution

Write $\vec v = 2\times10^5\,\hat i$ m/s (only x-component) and $\vec B = \hat i + 4\hat j + 3\hat k$ T.

Compute $\vec v\times\vec B = \begin{vmatrix}\hat i&\hat j&\hat k\\ v&0&0\\1&4&3\end{vmatrix} = \hat i(0\cdot3-0\cdot4)-\hat j(v\cdot3-0\cdot1)+\hat k(v\cdot4-0\cdot1) = (0,-3v,4v)$ .

Magnitude: $|\vec v\times\vec B| = v\sqrt{3^2+4^2} = v\times5 = 2\times10^5\times5 = 1\times10^6$ (in SI units, this is $\mathrm{m/s\cdot T}$ ).

Force magnitude: $F = |q|\,|\vec v\times\vec B| = 1.6\times10^{-19}\times10^6 = 1.6\times10^{-13}$ N.

This matches option (B).

Common Mistakes …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$Q. If the magnetic field is along the positive y-axis and the electron is moving along the positive x-axis, the direction of force on the electron is (A) along -y axis (B) along +y axis (C) along +z axis (D) along -z axis$
›Reveal solutionSolution
Using $\vec F = q\vec v\times\vec B$ with $v$ along $+x$ , $B$ along $+y$ , and the electron's negative charge, the force points along $-z$ .

Concept and Intuition

The magnetic (Lorentz) force on a moving charge is $\vec F = q\vec v \times \vec B$ . The cross product direction is found from the right-hand rule, and then the sign of the charge flips the direction if the charge is negative (as for an electron).

Step-by-Step Solution

$\vec v = v\hat x$ , $\vec B = B\hat y$ .

$\vec v \times \vec B = vB(\hat x \times \hat y) = vB\hat z$ . …
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Q.A charged particle of charge eee and mass mmm is moving in an electric field E⃗\vec{E}E and magnetic field B⃗\vec{B}B. Construct dimensionless quantities and quantities of dimension [T]−1[T]^{-1}[T]−1.

Why Magnetic Force Balance?

Step-by-step construction

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Q.A charged particle of charge $e$ and mass $m$ is moving in an electric field $\vec{E}$ and magnetic field $\vec{B}$ . Construct dimensionless quantities and quantities of dimension $[T]^{-1}$ .

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